Problem:
A circle of radius r is concentric with and outside a regular hexagon of side length 2. The probability that three entire sides of the hexagon are visible from a randomly chosen point on the circle is 1/2. What is r?
Answer Choices:
A. 22+23
B. 33+2
C. 26+3
D. 32+6
E. 62−3 Solution:
Place the hexagon in a coordinate plane with center at the origin O and vertex A at (2,0). Let B,C,D,E, and F be the other vertices in counterclockwise order.
Corresponding to each vertex of the hexagon, there is an arc on the circle from which only the two sides meeting at that vertex are visible. The given probability condition implies that those arcs have a combined degree measure of 180∘, so by symmetry each is 30∘. One such arc is centered at (r,0). Let P be the endpoint of this arc in the upper half-plane. Then ∠POA=15∘. Side BC is visible from points immediately above P, so P is collinear with B and C. Because the perpendicular distance from O to BC is 3, we have
Call the hexagon ABCDEF. Side AB is visible from point X if and only if X lies in the half-plane that is in the exterior of the hexagon and that is determined by the line AB. The region from which the three sides AB,BC, and CD are visible is the intersection of three such half-planes.
Let rays AB and DC intersect the circle at R and Q, respectively. Then QR is one of the six arcs of the circle from which three sides are visible. Symmetry implies that the six arcs are congruent, and because the given probability is 1/2, the measure of each arc is 30∘. Let O be the center of the hexagon and the circle. Then ∠QOR=30∘, so
